IMPULSE AND MOMENTUM 255
2. A uniform chain is hung from its upper end so that its lower end just touches an inelastic horizontal table, and then it is allowed to fall. Find the force which the table will experience at any instant during the fall of the chain.
The force is partly due to the weight of that part of the chain which is on the table at the instant considered and partly due to the rate at which the table is receiving momentum. Let x be the height of the upper end of the chain above the table, I the total length, and p the mass per unit length. Then pg (I — x) is the weight of that part of the chain which is on the table. On the other hand the momentum which the table receives in the interval of time dt is pv dt • v. Therefore the rate at which it receives momentum is pi>2, where v is the velocity of that part of the chain which is above the ground. This velocity is the same as that of-the upper end of the chain, therefore
„= V2g(l-x). Hence the total force is
F = p(l-x)g + p.'2g(l-x) = 3p(l-x)g.
DISCUSSION. — When x = Z, thai is, at the beginning of the motion, the force is zero. When x = 0, that is, at the end of the motion, it is 3 pig, or three times the weight of the chain. As soon as the entire chain comes to rest on the table the force equals the weight of the chain.
3. A spherical raindrop, descending by virtue of its weight, receives continuously, by precipitation of vapor, an accession of mass proportional to the surface. Find the velocity at any instant.
The external force acting upon the drop at any instant equals the rate at which its momentum changes, therefore
mg = - (mv), (1)
where m, the mass of the drop, is variable. Since the accession of mass is proportional to the surface the rate of change of radius of the drop will be constant. Let a be the radius of the drop when it begins to fall, r its radius at any later instant, and k the rate at which r increases. Then at .any instant . ,
m = T 17TT3